I'm not sure I'm following your steps. Why did you place an x infront in your first step?

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Please clarify.

Apr 17th 2007, 02:53 PM

ThePerfectHacker

Quote:

Originally Posted by phack

I'm not sure I'm following your steps. Why did you place an x infront in your first step?

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Please clarify.

Because,

ln(1+x)= x - x^2/2+x^3/3 -... for -1<x<=1

But you have,

ln(1+x^2)

Thus, instead of x you have x^2.

Apr 17th 2007, 08:33 PM

CaptainBlack

Quote:

Originally Posted by phack

I'm not sure I'm following your steps. Why did you place an x infront in your first step?

Because in your question you have an integral over x, but no x in the
integrand. Therefore the integral is x times whatever you had under
the integral sign (as this was independent of x). Therefore the first step
is the answer to the question you asked. After that I proceeded to answer
the question I assumed that you realy wanted to ask.

Quote:

And how did you get this?...

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

Please clarify.

As ImPerfectHacker says, I took a standard series representation of
log(1+u) and evaluated it at u=x^2. You can find this series either by
just knowing it, or taking the Taylor series of log(1+x) about 0, or by
integrating term by term the series expansion of 1/(1+x).